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TNPSC GROUP I PRELIMINARY EXAMINATION – 03.03.2019

APTITUDE & MENTAL ABILITY SOLUTIONS

1. In a right triangular ground, the sides adjacent to the right angle are 50 m and 80 m. Find the cost of cementing the ground at Rs.5/sq.m A. Rs.20,000 B. Rs.15,000 C. Rs.10,000 D. Rs.12,000 xU tpisahl;Lj;jply; nrq;Nfhz Kf;Nfhzk; tbtpy; cs;sJ. nrq;Nfhzj;ijj; jhq;Fk; gf;fq;fs; 50 kP> 80kP jplypy; rpnkz;l; G+r rJu kPl;lUf;F & 5 tpjk; MFk; nkhj;j nryitf; fhz;. A. ì 20,000 B. ì 15,000 C. ì 10,000 D. ì 12,500

2. The average of 5 numbers is 180. If one of the numbers is removed then the average becomes 155. Find the removed number? A. 240 B. 280 C. 320 D. 360 5 vz;fspy; ruhrhp 180 mtw;wpy; xU vz;iz ePf;fpdhy; ruhrhp 155 vd khWfpwJ vdpy; ePf;fg;gl;l vz;izf; fhz;f. A. 240 B. 280 C. 320 D. 360 Explanation:

Sum of 5 Numbers = 180 5 900 Sum of 4 Numbers = 155 4 620

Removed Number = 900 – 620 = 280

3. If LCM of “a” and “b” is a and LCM of “b” and “c” is b, then what is the LCM

of “c” and “a”

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A. a B. b C. c D. abc a kw;Wk; b d; kP.ngh. klq;F a vdTk; b kw;Wk; c d; kP.ngh. klq;F b vdTk; ,Ue;jhy; c kw;Wk; ad; kP.ngh.k vd;d? A. a B. b C. c D. abc Explanation:

Let, a = 10 b = 5 c = 1

LCM (a, b) = a → LCM (10, 5) = 10 LCM (b, c) = b → LCM (5, 1) = 5 LCM (c, a) = ? → LCM (1, 10) = 10 ∴ LCM (c, a) = a

4. Which one is correct?

A. 3 13 15 1 1

5 40 80 20 2

B.

3 1 13 15 1

5 20 40 80 2

C. 15 13 1 3 1

80 40 20 5 2

D.

1 3 1 13 15

2 5 20 40 80

fPo;fz;ltw;wpy; vJ rhpahdJ?

A. 3 13 15 1 1

5 40 80 20 2

B.

3 1 13 15 1

5 20 40 80 2

C. 15 13 1 3 1

80 40 20 5 2

D.

1 3 1 13 15

2 5 20 40 80

Explanation:

3 16 48

5 16 80

13 2 26

40 2 80

15

80

1 4 4

20 4 80

1 40 40

2 40 80

48 26 15 4 40

Ans: A. 3 13 15 1 1

5 40 80 20 2

5. Calculate the compound interest on Rs.9,000 in 2 years when the rate of interest for successive years are 10% and 12% respectively A. Rs.1,188 B. 2,088 C. 4,396 D. 2,596 xU njhiff;F mLj;jLj;j Mz;LfSf;F KiwNa 10% kw;Wk; 12% tl;b tPjj;jpy; 9,000 f;F 2 Mz;LfSf;F njhlh; tl;b vt;tsT? A. ì 1,188 B. ì 2,088 C. ì 4,396 D. ì 2,596 Explanation:

110 112. 9000 9000 11088 9000 2088

100 100C I

6. A certain sum of money in simple interest scheme amounts to Rs.8,880 in 6

years and Rs.7,920 in 4 years respectively. Find the principal and rate percent A. Principal = 6,000, rate = 8% B. Principal = 6,600, rate = 8% C. Principal = 6,000, rate = 7% D. Principal = 6,600, rate = 7%

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xU Fwpg;gpl;l mryhdJ jdp tl;b tPjj;jpy; 6 Mz;Lfspy; 8,880 MfTk; 4 Mz;Lfspy; 7,920 MfTk; khWfpwJ vdpy; mry; kw;Wk; tl;b tPjj;ijf; fhz;f. A. mry; = 6,000 tl;b tPjk; = 8% B. mry; = 6,000 tl;b tPjk; = 8% C. mry; = 6,000 tl;b tPjk; = 7% D. mry; = 6,000 tl;b tPjk; = 7% Explanation:

7. A fair die is rolled. Find the probability of getting a prime factor of 6

A. 2

3 B.

1

3 C.

5

6 D.

1

2

xU rPuhd gfil xU Kiw cUl;lg;gLfpwJ. mjpy; 6d; gfh fhuzpfs; fpilf;f epfo;jfT ahJ?

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A. 2

3 B.

1

3 C.

5

6 D.

1

2 Explanation:

Total No.of Events = 6 getting a prime factor = 2, 3

Required Probability = 2 1

6 3

8. Find the median of the data

12, 14, 25, 23, 18, 17, 24, 20 A. 23 B. 18 C. 17 D. 19

12, 14, 25, 23, 18, 17, 24, 20 vd;w tptuq;fspd; ,ilepiyia fhz;f. A. 23 B. 18 C. 17 D. 19 Explanation:

12, 14, 17, 18, 20, 23, 24, 25

Median = 18 20 38

192 2

9. In a two digit number, the digit in the unit place is twice of the digit in the

tenth place. If the digits are reversed, the new number is 27 more than the given number. find the number A. 63 B. 36 C. 93 D. 39 Xh; <hpyf;f vz;zpy;> xd;whk; ,l ,yf;f vz;> gj;jhk; ,l ,yf;f vz;izg;Nghy; ,U klq;fhf cs;sJ. ,yf;fq;fs; ,lk; khwpdhy; fpilf;Fk; Gjpa vz; nfhLf;fg;gl;l vz;iztpl 27 mjpfk; vdpy; nfhLf;fg;gl;l <hpyf;f vz;izf; fz;Lgpbf;f. A. 63 B. 36 C. 93 D. 39 Explanation:

Number by x x=units digit y=tens digit x=2y digits reversed: yx xy = yx +27 10x+y=10y+x+27 20y + y = 10y+2y+27 9y=27 y=3 x=2y=6 number:36 or Unit place = 2 Tens place Solve from option easier. Possible Number = 36 Old Number = 36 New Number = 63 [36+27]

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10. The population of a village is 32,000. 40% of them are men. 25% of them are

women and the rest are children. Find the number of men and children A. 12200 men, 11800 children B. 12800 men, 11200 children C. 12220 men, 1220 children D. 12200 men, 11200 children xU fpuhkj;jpd; kf;fs; njhif 32>000 cs;sdh;. mth;fspy; 40% Mz;fs;> 25% ngz;fs;> kPjk; cs;Nshh; Foe;ijfs;. Mz;fs; kw;Wk; Foe;ijfspd; vz;zpf;ifiaf; fhz;f. A. 12200 Mz;fs;> 11800 Foe;ijfs; B. 12800 Mz;fs;> 11200 Foe;ijfs; C. 12220 Mz;fs;> 12200 Foe;ijfs; D. 12200 Mz;fs;> 11200 Foe;ijfs; Explanation:

40No.of Men=32000 1,28,000

100

35No.of Children=32000 11,200

100

11. Using clay, a student made a right circular cone of height 48 cm and base

radius 12 cm. another student reshapes it in the form of sphere. Find the radius of the sphere. A. 12 cm B. 15 cm C. 9 cm D. 14 cm fspkz;izg; gad;gLj;jp xU khztd; 48 nr.kP cauKk; 12 nr.kP MuKk; nfhz;l Neh; tl;l jpz;kf; $k;igr; nra;jhh;. mf;$k;ig kw;nwhU khzth; xU jpz;kf; Nfhskhf khw;wpdhh;. mt;thW khw;wg;gl;l Gjpa Nfhsj;jpd; Muj;ijf; fhz;f. A. 12cm B. 15 cm C. 9 Cm D. 14 Cm Explanation: Given that radius of the cone, r =12 cm and height of the cone h = 48 cm We wish to find the radius, R of the sphere Now Volume of the sphere = Volume of the cone

34

3R = 21

3r h

3R = 21

4r h

=

112 12 48

4

3R = 12 12 12

R = 12 cm

12. The radius of a spherical balloon increases from 3 cm to 9 cm as air is being pumped into it. Find the ratio of volumes of the balloon in the two cases. A. 1 : 3 B. 1 : 9 C. 1 : 27 D. 1 : 8 3 nr.kP Muk; nfhz;l Nfhs tbt gY}dpy; fhw;W nrYj;jg;gLk; NghJ mjd; Muk; 9 nr.kP Mf> mjpfhpj;jhy;> mt;tpU epiyfspy; gY}dpd; fd msTfspd; tpfpjj;ijf; fhz;f.

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A. 1 : 3 B. 1 : 9 C. 1 : 27 D. 1 : 8 Explanation:

3 34 4

3 : 9 3 3 3 : 9 9 9 1: 273 3

13. If the circle with diameter 4 cm is inside of the circle with radius 10 cm. Then

the Maximum possible distance between the centres is A. 6 B. 7 C. 8 d. 9 4 nr.kP tpl;lj;ij nfhz;l tl;l 10 nr.kP MuKila tl;lj;jpDs; cs;s NghJ ,U ikaq;fSf;F ,ilNa ,Uf;f ,aYk; mjpfgl;r J}uk; vd;d? A. 6 B. 7 C. 8 D. 9 Explanation:

R = 10 cm r=2 cm Distance between the centres = R – r = 10 – 2 = 8 cm

14. A person crosses 600 m long street in 5 minute. What is his speed in km per

hour A. 3.6 B. 7.2 C. 8.4 D. 10 xU egh; 600 kPl;lh; ePsKs;s xU njUit 5 epkplq;fspy; flf;fpwhh; vdpy; mthpd; Ntfj;ij fp.kP / kzpapy; $W. A. 3.6 B. 7.2 C. 8.4 D. 10 Explanation:

D 600m 18 36S= 2 7.2kmph

T 300s 5 5

15. A man travelled 2

11 of his journey by coach,

17

22 by rail and walked the

remaining 1 km. How far did he go? A. 22 km B. 20 km C. 33 km D. 27 km xUth; jdJ gaz J}uj;jpy; gq;if Nfhr;R tz;bapYk; gq;if uapy; tz;bapYk; kPjKs;s 1 fpNyhkPl;liu ele;Jk; flf;fpwhh; vdpy; gaz J}uk; vd;d? A. 22 fp.kP B. 20 fp.kP C. 33 fp.kP D. 27 fp.kP Explanation:

2 17 4 17 21

11 22 22 22

Remaining Distance 22-21 = 1unit = 1km Total Distance Travelled = 22 1=22 km

16. A ramp for unloading a moving truck has an angle of elevation of 30°. If the

top of the ramp is 0.9 m above the ground level then find the length of the ramp A. 2 m B. 1.5 m C. 1.8 m D. 1.6 m

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xU Rik Ch;jpapUe;J Rikia ,wf;f VJthf 30° Vw;wf; Nfhzj;jpy; xU rha;Tj; jsk; cs;sJ. rha;T jsj;jpd; cr;rp jiuapypUe;J 0.9 kP cauj;jpy; cs;sJ vdpy;> rha;Tj;jsj;jpd; ePsk; ahJ? A. 2 kP B. 1.5 kP C. 1.8 kP D. 1.6 kP Explanation:

17. Simplify: 2 2

2

25 ( 5)

3 9

x x

x x

A. ( 5)( 3)

( 3)

x x

x

B.

( 5)( 3)

( 5)

x x

x

C. ( 5)( 3)

( 3)

x x

x

D.

( 5)( 3)

( 5)

x x

x

Explanation:

5 5 3 3 5 3

3 5 5 5

x x x x x x

x x x x

18. Simplify : 2 1 2 5

7 4 3 6

A. 168

125 B.

125

160 C.

125

168 D.

160

125

RUf;Ff: 2 1 2 5

7 4 3 6

A. 168

125 B.

125

160 C.

125

168 D.

160

125

Explanation:

1 3 3

4 2 8

Required: 2 3 5 2 9 20 2 11 2 11 48 77 125

7 8 6 7 24 7 24 7 24 168 168

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19. Find the greatest number of four digits exactly divisible by 15, 21 and 27 A. 9999 B. 9450 C. 9845 D. 8505 15> 21 kw;Wk; 27 Mfpa %d;W vz;fshy; kPjpapd;wp tFgLk; kpfg;nghpa ehd;F ,yf;f ,ay; vz; vJ? A. 9999 B. 9450 C. 9845 D. 8505 Explanation:

Largest 4 digit Number = 9999 LCM of(15, 21 and 27) =945 9999 divisible by 945 get the remainder 549.

Required Number = 9999 – 549 = 9450

20. The GCD and LCM of two polynomials are x + 1 and x4 – 1 respectively. If one of the polynomials is x2 + 1, find other one. A. x3 – 1 B. (x + 1)(x2-1) C. x2 + x – 1 D. x2 – x + 1 ,U gy;YWg;Gf; Nfhitfspd; kP.ngh.t. kw;Wk; kP.ngh.k KiwNa x + 1 kw;Wk; x4 – 1, NkYk; xU gy;YWg;Gf;Nfhit x2 + 1 vdpy;> kw;nwhd;iw fhz;f. A. x3 – 1 B. (x+1) (x2-1) C. x2+x-1 D. x2 – x + 1 Explanation:

Other Number =

2 2

2

2

1 1 1LCM×HCF1 1

Given Number 1

x x xx x

x

21. Find the Highest common Factor of 2 3 3 2 2 2 4 34 ,8 ,16p q r p q r p q r

A. 2 4 34p q r B. 2 24p q r C. 2 216p q r D. 2 4 316p q r

kPg;ngU nghJtFj;jp fhz;f. 4 p2p3r, 9p3p2r2, 16p2q4r3 A. 4p2q4r3 B. 4p2q2r C. 16p2q2r D. 16p2q4r3 Explanation: HCF(4, 8, 16) = 4 HCF of 2 3 3 2 2 2 4 3, ,p q r p q r p q r = 2 2p q r Required HCF= 2 24p q r

22. What is the half of the area of the triangle whose vertices are (1,1) (3,1), (1,3)?

xU Kf;Nfhzj;jpd; Kidfs; (1,1), (3,1), (1,3) vdpy; mjd; gug;gstpy; ghjp fhz;f. A. 1 B. 2 C. 4 D. 5 Explanation:

1 2 3 2 3 1 3 1 2

1

2

11 1 3 3 3 1 1 1 1

2

12 6 0

2

14 2 .

2

x y y x y y x y y

sq unit

Half of Area = 1 sq.unit

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23. A silver wire when bent in the form of a square encloses an area of 484 sqm. If the same wire is bent in the form of a circle, then find the diameter of the

circle (use 22

7 )

A. 14m B. 28 m C. 24 m D. 7 m xU nts;spf; fk;gp tisf;fg;gl;L rJukhf khw;Wk; NghJ. mjdhy; milgLk; gFjpapd; gug;G 484 r.kP mNj nts;spf; fk;gpia tl;lkhf tisf;fg;gLfpwJ vdpy; tl;lj;jpd; tpl;lk; vd;d? ( vdf;nfhs;f) A. 14 kP B. 28 kP C. 24 kP D. 7 kP Explanation:

a2 = 484 ∴ a = 22 Perimeter of square = Circumference of a circle

4 x a = 2πr

4 x 22 = 22

27

r

r = 14 m Diametre = 28 m

24. In how much time will a sum of Rs.1600 amount to Rs.1852.20 at 5% per

annum compound interest A. 2 years B. 3 years C. 4 years D. 5 years ì 1600 MdJ 5% Mz;L $l;L tl;b tPjk; nfhz;L vj;jid Mz;Lfspy; ì 1852.20 MFk;? A. 2 Mz;Lfs; B. 3 Mz;Lfs; C. 4 Mz;Lfs; D. 5 Mz;Lfs;

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25. The standard deviation of 10 values is 3. If each value in increase by 4. Find the variance of the new set of value

A. 30 B. 12 C. 9 D. 3 10 kjpg;Gfspd; jpl;l tpyf;fk; 3. xt;nthU kjpg;GlDk; 4-If; $l;l fpilf;Fk; Gjpa jpl;l tpyf;fj;jpd; tpyf;f th;f;f ruhrhp ahJ?

A. 30 B. 12 C. 9 D. D. 3 Explanation:

In S.D. same number added, S.D. result does not change. New = 3 Variance = 2 23 9

26. Find the next term of this sequence 11, 13, 17, 19, 23 __________

A. 25 B. 27 C. 29 D. 31 11> 13> 17> 19> 23> _____________ vd;w njhlh; thpirapy; mLj;J tUk; cWg;G vJ? A. 25 B. 27 C. 29 D. 31 Explanation: First Method: Prime Number Series: 11, 13, 17, 19, 23, 29. Second Method: 11+2=13, 13+4=17, 17+2=19, 19+4=23, 23+2=25

27. Find the mean of 2, 4, 6, 8, 10, 12, 14, 16 A. 10 B. 9 C. 12 D. 14 $l;Lr; ruhrhp fhzTk;. 2> 4> 6> 8> 10> 12> 14> 16 A. 10 B. 9 C. 12 D. 14 Explanation:

Mean = 2 4 6 8 10 12 14 16 72

98 8

28. The exterior angles of a pentagon are in the ratio 6 : 3 : 4 : 3 : 2. Find all its

interior angles. A. 60°, 120°, 80°, 160°, 120° B. 80°, 110°, 150°, 120°, 80° C. 100°, 170°, 160°, 40°, 70° D. 60°, 120°, 100°, 120°, 140° xU Iq;Nfhzj;jpd; ntspf;Nfhzq;fs; 6 : 3 : 4 : 3 : 2 vd;w tpfpjj;jpy; cs;sd vdpy; mjd; cl;Nfhzq;fspd; msTfisf; fhz;f. A. 60°, 120°, 80°, 160°, 120° B. 80°, 110°, 150°, 120°, 80° C. 100°, 170°, 160°, 40°, 70° D. 60°, 120°, 100°, 120°, 140° Explanation:

Exterior Angle of a pentagon = 360°

360 360

206 3 4 3 2 18

∴ 120°, 60°, 80°, 60°, 40° Interior Angle = 180° - Exterior Angle

∴ 60°, 120°, 100°, 120°, 140°

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29. Which one of the following cannot be the sides of a triangle? A. 4, 5, 6 B. 3, 4, 5 c. 2, 3, 4 D. 1, 2, 3 gpd;tUtdtw;Ws; vit Kf;Nfhzj;jpd; gf;fq;fshf ,Uf;f KbahJ? A. 4> 5> 6 B. 3> 4> 4 C. 2> 3> 4 D. 1> 2> 3 Explanation:

Largest side < Sum of other two sides In above rule, Option D is not satisfied. 3 = 3 30. Without using logarithm table find approximate value for 10log 2

A. 0.2401 B. 0.3 C. 0.2802 D. 1.414 klf;if ml;ltiziag; gad;gLj;jhky; -d; Njhuha kjpg;G fhz;f. A. 0.2401 B. 0.3 C. 0.3802 D. 1.414 Explanation:

31. Pocket money received by 7 students is given below. ` 42, ` 22, ` 40, ` 26, ` 23, ` 28, ` 43 Find the median. A. ` 26 B. ` 23 C. ` 28 D. ` 22 VO khzth;fs; jpdr; nrytpw;fhf thq;fpa gzk; fPNo nfhLf;fg;gl;Ls;sJ.

` 42, ` 22, ` 40, ` 26, ` 23, ` 28, ` 43 ,tw;wpd; ,ilepiyasitf; fhz;f.

A. ` 26 B. ` 23 C. ` 28 D. ` 22 Explanation:

22, 23, 26, 28, 40, 42, 43 Median = 28

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32. The Range of the first 10 prime numbers is

A. 28 B. 26 C. 29 D. 27 Kjy; 10 gfh vz;fspd; tPr;R.

A. 28 B. 26 C. 29 D. 27 Explanation:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29 Range = L – S = 29 – 2 = 27

33. The students of a class donated ` 4624 for Chief Minister‟s State Relief Fund.

Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

A. 64 B. 68 C. 62 D. 78 xU tFg;gpy; cs;s khzth;fs; &. 4624-ia Kjyikr;rhpd; khepy epthuz epjpahf nrYj;jpdh;. xt;nthU khztUk; jdJ gq;fhf jdJ tFg;gpy; cs;s khzth;fspd; vz;zpf;iff;F rkkhd njhifia nfhLj;jdh; vdpy; tFg;gpy; cs;s khzth;fs; vj;jid Ngh;? A. 64 B. 68 C. 62 D. 78

Explanation: No. of Students = x Each students donated = x Rs. x x x = 4624

x = 4624 = 68

34. Simplify the following:

3 8

2 2

x

x x

A. x2 – 2x + 4 B. x2 + 2x + 4

C. x2 – 2x – 4 D. x2 + 2x – 4 RUf;Ff.

3 8

2 2

x

x x

A. x2 – 2x + 4 B. x2 + 2x + 4 C. x2 – 2x – 4 D. x2 + 2x – 4

Explanation:

3 3 38 2

2 2 2

x x

x x x

= 22 2 4

( 2)

x x x

x

= x2 +2x + 4

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35. Simplify 2

1 1

1 1 1

m

m m m

A. m2 B. 2

2

1m

m

C.

2

1

m D.

2

2 1

m

m

RUf;Ff.

2

1 1

1 1 1

m

m m m

A. m2 B. 2

2

1m

m

C.

2

1

m D.

2

2 1

m

m

Explanation:

= 2

1 1

1 1

m

m m

= 2 2

2 2 2

1 1 11

1 1 1

m m

m m m

36. If 3a + 4b = 22, 8a – 5b = -4, then the value of (a+b)2 = ? A. 36 B. 72 C. 25 D. 49 3a + 4b = 22, 8a – 5b = -4 vdpy; (a+b)2 d; kjpg;G ahJ? A. 36 B. 72 C. 25 D. 49 Explanation:

Solving Equation a = 2, b = 4 (a + b)2 = (2 + 4)2 = (6)2 = 36

37. If 123 represents “GOD” 456 represents “CAT”, Then “DOG EAT EGG” may

be represented by A. 321 456 411 B. 321 756 811 C. 321 856 911 D. 321 756 711

123> vd;gJ “GOD” vd;gijAk; 456 vd;gJ “CAT” vd;gijAk; Fwpj;jhy; “DOG EAT EGG” vd;gJ vjdhy; Fwpf;fg;glyhk;? A. 321 456 411 B. 321 756 811 C. 321 856 911 D. 321 756 711

Explanation: D O G E A T E G G 3 2 1 7 5 6 7 1 1

„E‟ Value Except 1, 2, 3, 4, 5, 6 ∴ E = 7 Ans: D. 321 756 711

38. Find the number of 3 digit natural numbers which are divisible by 6 A. 151 B. 150 C. 152 D. 14 6-My; tFglf; $ba 3 ,yf;f ,ay; vz;fspd; vz;zpf;ifia fhz;f. A. 151 B. 150 C. 152 D. 149 Explanation:

102, 108, ………. 996

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n = 996 102

1 16

l a

d

= 894

16

= 149 + 1 = 150

39. What is the probability that a leap year selected at random will contain 53 Sundays?

A. 2

7 B.

3

7 C.

4

7 D.

5

7

rk tha;g;G Kiwapy; Njh;e;njLf;fg;gLk; yPg; tUlk; 53 Qhapw;Wf; fpoikfis nfhz;L ,Uf;f epfo;jfT ahJ?

A. 2

7 B.

3

7 C.

4

7 D.

5

7

Explanation: Now 52 weeks contain 52 Sundays and the remaining two days will be one of the following seven possibilities. (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thur), (Thur, Fri), (Fri, Sat) and (Sat, Sun). The probability of getting 53 Sundays in a leap year is same as the probability of getting a Friday in the above seven possibilities. Here S=(Sun,Mon),(Mon,Tue),(Tue,Wed),(Wed,Thur),(Thur,Fri),(Fri,Sat),(Sat,Sun),. Then n(S) = 7. Let A be the event of getting one Sunday in the remaining two days.

40. Shyam‟s monthly income is Rs. 12,000. He saves Rs. 1200. Find the percent of his savings and his expenditure.

A. 10%, 80% B. 10%, 90% C. 80%, 10% D. 90%, 10% ~pahkpd; khj tUkhdk; &. 12>000. mth; Nrkpf;Fk; njhif &. 1>200. mthpd; Nrkpg;G> nryT Mfpatw;wpd; rjtPjj;ijf; fhz;f. A. 10%, 80% B. 10%, 90% C. 80%, 10% D. 90%, 10%

Explanation:

Saving = 1200

10 10%12000

Expenditure = 90%

41. Due to increase of 30% in the price of a colour TV the sale is reduced by 40%. What will be the percentage change in income?

A. 10% increase B. 10% decrease C. 35% decrease D. 22% decrease

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xU tz;z njhiyf;fhl;rp ngl;bapd; tpiyapy; 30% mjpfhpj;jjhy; 40% tpw;gid Fiwe;jJ vdpy; tUthapy; Vw;gLk; khw;wj;ij rjtPjj;jpy; $W. A. 10% mjpfhpg;G B. 10% FiwT C. 35% FiwT D.22% FiwT

Explanation:

= 30 – 40 - 30 40

100

= 30 – 40 – 12 = - 22%

42. Find the least number which when divided by 24, 32 and 36 leaves the remainders 19, 27 and 31 respectively.

A. 280 B. 290 C. 283 D. 300 xU vz;iz 24> 32 kw;Wk; 36My; tFf;Fk; NghJ KiwNa 19> 27 kw;Wk; 31 vd;gd kPjpahf fpilj;jhy; me;j vz;iz fhz;.

A. 280 B. 290 C. 283 D. 300 Explanation:

(24 – 19), (32 – 27), (36 – 31) = 5 Required Number = LCM (24, 32, 36) – 5 288 – 5 = 283.

43. The cost of levelling and turfing a square lawn at Rs. 2.50 per m2 is Rs.

13322.50. Find the cost of fencing if at Rs. 5 per metre. A. ` 1500 B. ` 1380 C. ` 1225 D. ` 1460 xU rJu tbt taiy kz; mbj;J rkg;gLj;j rJu kPl;lUf;F &. 2.50 tPjk; &. 13322.50 nrythfpwJ. vdpy; me;j taypw;F> Ntyp mikf;f kPl;lUf;F &. 5 tPjk; vt;tsT nryT MFk;?

A. ` 1500 B. ` 1380 C. ` 1225 D. ` 1460 Explanation:

a2 = 13322.5

2.5 = 5329 = (73)2

a = 73 Perimetre 4a = 4 x 73 = 292 Cost = 292 x 5 = 1460

44. The diameter of a semicircular grass plot is 70 m. Find the cost of fencing the

plot at ` 12 per metre. A. ` 7700 B. ` 840 C. ` 2160 D. ` 4320 miu tl;l tbtpyhd Gy;ntsp xd;wpd; tpl;lk; 70 m. mjw;Fr; Rw;W Ntyp mikf;f xU kPl;lUf;F &. 12 tPjk; vd;d nrythFk;?

A. &. 7700 B. &. 840 C. &. 2160 D. &. 4320 Explanation:

Circumference = πr + 2r = 36

7 x r

Required cost = 36

35 127

= 36 x 5 x 12 = 2160

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45. A circus tent is to be erected in the form of a cone surmounted on a cylinder.

The total height of the tent is 49 m diameter of the base is 42 m and height of the cylinder is 21 m. Find the cost of canvas needed to make the tent, if the cost of convas is ` 12.50/m2

A. ` 63552 B. ` 65352 C. ` 63525 D. ` 65532 xU rh;f;f]; $lhukhdJ cUisapd; kPJ $k;G ,ize;j tbtpy; mike;Js;sJ. $lhuj;jpd; nkhj;j cauk; 49 kP mjd; mbg;gf;fj;jpd; tpl;lk; 42 kP cUisg; ghfj;jpd; cauk; 21 kP NkYk; 1 r.kP. fpj;jhd; Jzpapd; tpiy &. 12.50 vdpy;> $lhuk; mikf;fj; Njitahd fpj;jhd; Jzpapd; tpiyiaf; fhz;f.

A. &. 63552 B. &. 65352 C. &. 63525 D. &. 65532

46. The average height of 10 students in a class was calculated as 150 cm. On verification it was found that one reading was wrongly recorded as 130 cm instead of 140 cm. Find the correct mean height.

A. 150 cm B. 152 cm C. 153 cm D. 151 cm xU tFg;gpy; cs;s 10 khzth;fspd; ruhrhp cauk; 150 nr.kP vdf; fzf;fplg;gl;lJ. rhpghh;f;Fk; NghJ 140 nr.kP vd;gij 130 nr.kP vd jtwhf vLj;Jf; nfhs;sg;gl;ljhf njhpate;jJ. rhpahd ruhrhp cauk; fhz;f. A. 150 nr.kP B. 152 nr.kP C. 153 nr.kP D. 151 nr.kP Explanation:

Difference = 140 – 130 = 10

Average Increased = 10

110

New Average = 150 + 1 = 151 cm

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47. What is Mean of all two digit Numbers? A. 54.5 B. 49.5 C. 45.5 D. 44.5 midj;J <hpyf;f vz;fspd; ruhrhp vd;d? A. 54.5 B. 49.5 C. 45.5 D. 44.5 Explanation:

Sum = 10 + 11 + 12 + …… + 99 = (1 + 2 + 3 + …….. + 99) – (1 + 2 + …… + 9)

= 99 100 9 10

2 2

= 4950 - 45 = 4905

Required Average = 4905

90 = 54.5

48. If ax2 – 28x + 49 = 0 has equal roots. Then the value of a is

A. 1 B. 2 C. 3 D. 4 ax2 – 28x + 49 = 0 vd;w rkd;ghl;bd; %yq;fs; rkk; vdpy; α d; kjpg;G A. 1 B. 2 C. 3 D. 4 Explanation:

b2 – 4ac = 0 b2 = 4ac

c = 2

4

b

a

a = 2 28 28

4 4 49

b

c

= 4

49. I have a box which has 3 green, 9 blue, 4 yellow, 8 orange coloured cubes in it.

a. What is the ratio of orange to yellow cubes? b. What is the ratio of green to blue cubes? c. How many different ratios can be formed, when you compare each colour

to any one of the other colours? (a) (b) (c)

A. 3:1 1:2 10 ratios B. 1:3 2:1 4 ratios C. 1:2 3:1 12 ratios D. 2:1 1:3 12 ratios vd;dplKs;s xU ngl;bapy; 3 gr;ir> 9 ePyk;> 4 kQ;rs;> 8 MuQ;R vd 24 tz;zf; fdr; rJuq;fs; cs;sd. vdpy; a. MuQ;R kw;Wk; kQ;rs; fdr;ruq;fspd; tpfpjk; vd;d? b. gr;ir kw;Wk; ePyk; fdr;rJuq;fspd; tpfpjk; vd;d? c. xU tz;zj;ij kw;w tz;zq;fNshL xg;gpl;L vj;jid tpfpjq;fs; fhzyhk;?

(a) (b) (c) A. 3:1 1:2 10 ratios B. 1:3 2:1 4 ratios

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C. 1:2 3:1 12 ratios D. 2:1 1:3 12 ratios Explanation:

(a) Orange : Yellow 8 : 4 2 : 1

(b) Green : Blue 3 : 9 1 : 3

(c) Required No. of Ratios = 3 + 3 + 3 + 3 = 12 (Each colour compare with Remaining three)

50. A troop has provisions for 276 soldiers for 20 days. How many soldiers leave

the troop so that the provisions may last for 46 days. A. 136 B. 156 C. 146 D. 164 276 tPuh;fs; cs;s xU gl;lhsj;jpy; 20 ehl;fSf;Fj; Njitahd rikay; nghUs;fs; cs;sJ> me;jg; nghUs;fs; 46 ehl;fSf;F ePbf;f Ntz;Lnkdpy; vj;jid tPuh;fs; ,e;jg;gl;lhsj;ij tpl;Lr; nry;y Ntz;Lk;? A. 136 B. 156 C. 146 D. 164 Explanation:

No. of soldier = 276 20

46

= 120

No. of soldiers left = 276 – 120 = 156